Problem: A curve is defined by the parametric equations $x=t^2-16$ and $y=t^4+3t$. What is $\dfrac{d^2y}{dx^2}$ in terms of $t$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $2-\dfrac{3}{4t^3}$ (Choice B) B $\dfrac{2t}{4t^3+3}$ (Choice C) C $t+\dfrac{3}{4t^2}$ (Choice D) D $2t^2+\dfrac{3}{2t}$
Explanation: We are asked to find the second derivative of a parametric function. Recall that the first derivative of a function defined parametrically by the equations $x=u(t)$ and $y=v(t)$ is found with the following rule: $\dfrac{dy}{dx}=\dfrac{\left(\dfrac{dy}{dt}\right)}{\left(\dfrac{dx}{dt}\right)}=\dfrac{v'(t)}{u'(t)}$ Then, the second derivative is found with this following rule: $\dfrac{d^2y}{dx^2}=\dfrac{\dfrac{d}{dt}\left(\dfrac{dy}{dx}\right)}{\left(\dfrac{dx}{dt}\right)}=\dfrac{\dfrac{d}{dt}\left(\dfrac{v'(t)}{u'(t)}\right)}{u'(t)}$ Let's start by finding $\dfrac{dy}{dx}$. $\dfrac{dy}{dx}=2t^2+\dfrac{3}{2t}$ Now we can find $\dfrac{d^2y}{dx^2}$. $\begin{aligned} \dfrac{d^2y}{dx^2}&=\dfrac{\dfrac{d}{dt}\left(\dfrac{dy}{dx}\right)}{\left(\dfrac{dx}{dt}\right)} \\\\ &=\dfrac{\dfrac{d}{dt}\left(2t^2+\dfrac{3}{2t}\right)}{\dfrac{d}{dt}(t^2-16)} \\\\ &=\dfrac{4t-\dfrac{3}{2t^2}}{2t} \\\\ &=2-\dfrac{3}{4t^3} \end{aligned}$ In conclusion, $\dfrac{d^2y}{dx^2}=2-\dfrac{3}{4t^3}$.